I think he means the "fundamental solution" of the differential operator.
Daniel.

Actually, I would argue that that doesn't make sense either- an operator is not an equation. An equation may have a solution, but not the operator!

asdf1 said:

for example:
(D^2-d-2)y=0
if you solve D, which is D=2,-1

Okay, I can understand that, although your terminology is still odd!
I presume you are referring to the differential equation:
(D2- D- 2)y= 0 where "D" is the differential operator d/dx (or d/dt). D2- D- 2 would also be a linear differential operator.

However, solving the equation D2- D- 2= 0 to get D= 2 or D= -1 is what mathematicians call "abuse of terminology". If you intend D to be "d/dx", it clearly doesn't make sense to turn around and say that D= 2!
It is, though, convenient shorthand and we do it all the time. It is convenient shorthand for the "characteristic equation". If we were to look for a solution of the form [itex]y= e^{\lambda x}[/itex], putting that into the equation would give [itex]\lambda^2 e^{\lambda x}- \lambda e^{\lambda x}- 2e^{\lambda x}= 0[/itex] or [itex]\left(\lambda^2- \lambda- 2\right)e^{\lambda x}= 0[/itex]. Since [itex]y= e^{\lambda x}[/itex] is never 0, we must have [itex]\lambda^2- \lambda- 2= 0[/itex], the characteristic equation.
That "looks like" the original equation, especially in operator notation (which is the main reason for using it) because of that very nice property of exponentials:
[tex]\frac{d^n(e^{\lambda x}}{dx^n}= \lambda^n e^{\lambda x}[/tex]

Caution!! This is only true for a very limited (though important) class of differential equations: linear equations with constant coefficients.

sorry, i didn't make myself clear~
for the example above, usually the normal way is to suppose that
y=ce^(namda)x
and find namda
but the answer to namda is the same as finding the answer to the differential operator~